Projectile hits rod hanging from pivot.

AI Thread Summary
A thin, uniform bar weighing 90N is hanging from a ceiling pivot and is struck by a 3kg ball traveling horizontally at 10 m/s, which rebounds at 6 m/s. The discussion focuses on calculating the angular momentum before and after the collision, using the moment of inertia for both the ball and the rod. The initial calculations yield an angular velocity of 3.92 rad/s, while the expected answer is 5.88 rad/s, indicating an error in the moment of inertia formula. Participants clarify the correct syntax for expressing equations and emphasize the importance of accurately deriving moments of inertia. The conversation highlights common pitfalls in physics calculations and the need for precise notation.
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Homework Statement


A thin, uniform bar, 2, long and weighing 90N is hanging vertically from the ceiling by a frictionless pivot. It is struck by a small 3kg ball, 1.5m below the ceiling, initially traveling horizontally at 10 m/s. The ball rebounds in the opposite direction with a speed of 6 m/s.

Homework Equations



Lbefore = Lafter

L = Iω

Irod = \frac{1}{2}MR2

Ipoint = MR2

The Attempt at a Solution



At the point of impact the ball can be thought of as a particle in circular motion about the pivot with radius 1.5m and tangential velocity 10m/s.
The change in velocity is 16m/s. So the effective change in the angular momentum of the ball is

ΔLball = Iball Δωball = \frac{I<sub>ball</sub>Δv}{R<sub>ball</sub>}

This is equal to the change in the angular momentum of the rod (opposite direction):

ΔLrod = Irod Δωrod = \frac{I<sub>ball</sub>Δv}{R<sub>ball</sub>}

Insert values for I and rearrange :

Δωrod = (2*mball*rball*Δvball) / (mrod*rball2)

This gives ω = 3.92 rad/s , the given answer is 5.88 rad/s.
 
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I obviously used the fraction syntax incorrectly, please let me know what I did wrong. Thanks
 
BOYLANATOR said:

Homework Statement


A thin, uniform bar, 2, long and weighing 90N is hanging vertically from the ceiling by a frictionless pivot. It is struck by a small 3kg ball, 1.5m below the ceiling, initially traveling horizontally at 10 m/s. The ball rebounds in the opposite direction with a speed of 6 m/s.


Homework Equations



Lbefore = Lafter

L = Iω

Irod = \frac{1}{2}MR2

The formula for the moment of inertia of the rod is not correct.


ehild
 
BOYLANATOR said:
I obviously used the fraction syntax incorrectly, please let me know what I did wrong
itex gives up if you put non-itex codes like [s u b] inside. Use ^ for sup and _ for sub.
 
As haruspex said, _ for sub, but enclose subscript between curly thingies {} :smile:

ΔL_{ball} = I_{ball} Δω_{ball} = \frac{I_{ball}Δv}{R_{ball}}

written as

ΔL_{ball} = I_{ball} Δω_{ball} = \frac{I_{ball}Δv}{R_{ball}}

ehild
 
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Ah yes. I should stick to deriving the moments of inertia, my memory doesn't serve me well. Thanks.
 
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